A pair of vertical angles has measures (2y + 5) and (4y) What is the value of y?

Mathematics

1 answer:

iogann1982 [59]3 years ago

8 0

Answer:

Vertical angles are angles that are opposite each other when two lines intersect each other. And they are always equal.

==> 2y +5 = 4y

==> 4y - 2y = 5

==> 2y = 5

==> $y = \frac{5}{2}$

Option D

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What is the remainder of (4x2 + 7x-1)= (4 + x)? A. -9x – 1 B.23x – 1 C.35 D.-37

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$x=\frac{-3+\sqrt{29}}{4},\:x=-\frac{3+\sqrt{29}}{4}\\$

Step-by-step explanation:

$\left(4x^2+7x-1\right)=\left(4+x\right)\\\mathrm{Refine}\\4x^2+7x-1=4+x\\\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}\\4x^2+7x-1-x=4+x-x\\Simplify\\4x^2+6x-1=4\\\mathrm{Subtract\:}4\mathrm{\:from\:both\:sides}\\4x^2+6x-1-4=4-4\\\mathrm{Simplify}\\4x^2+6x-5=0\\\mathrm{For\:}\quad a=4,\:b=6,\:c=-5:\\\quad x_{1,\:2}=\frac{-6\pm \sqrt{6^2-4\times \:4\left(-5\right)}}{2\times \:4}$

$\frac{-6+\sqrt{6^2-4\times \:4\left(-5\right)}}{2\times \:4}\\=\frac{-6+\sqrt{6^2+4\times \:4\times \:5}}{2\times \:4}\\=\frac{-6+\sqrt{116}}{2\times \:4}\\=\frac{-6+\sqrt{116}}{8}\\\\Let\: simplify\: ; -6+2\sqrt{29}\\=-2\times \:3+2\sqrt{29}\\=2\left(-3+\sqrt{29}\right)\\=\frac{2\left(-3+\sqrt{29}\right)}{8}\\=\frac{-3+\sqrt{29}}{4}\\$

$\frac{-6-\sqrt{6^2-4\times \:4\left(-5\right)}}{2\times \:4}\\\\=\frac{-6-\sqrt{6^2+4\times \:4\times \:5}}{2\times \:4}\\\\=\frac{-6-\sqrt{116}}{2\times \:4}\\\\=\frac{-6-2\sqrt{29}}{8}\\\\=-\frac{2\left(3+\sqrt{29}\right)}{8}\\\\=-\frac{3+\sqrt{29}}{4}\\\\\\x=\frac{-3+\sqrt{29}}{4},\:x=-\frac{3+\sqrt{29}}{4}$